How many mL of a 2.00 M NaBr solution are needed to make 200.0 mL of 0.500M NaBr?

2 Answers
May 13, 2018

Approx. 50.0*mL....

Explanation:

"Concentration"="Number of moles"/"Volume of solution"...and so we use this quotient appropriately to find the THIRD value if the other two values are specified.

We want a 200*mL volume of 0.500*mol*L^-1 NaBr...

n_"sodium bromide"=200xx10^-3*Lxx0.500*mol*L^-1=0.10*mol.

And 2.00*mol*L^-1 NaBr(aq) is available...so....

(0.10*mol)/(2.00*mol*L^-1)xx1000*mL*L^-1=50.0*mL.

May 13, 2018

"50.0 mL" of the "2.00 M NaBr" solution is needed to make "200.0 mL" of a "0.500 M NaBr" solution.

Explanation:

Use the dilution equation:

C_1V_1=C_2V_2,

where:

C_1 is the initial concentration, V_1 is the initial volume, C_2 is the final concentration, and V_2 is the final volume.

Known

C_1="2.00 M"

C_2="0.500 M"

V_2=200.0"mL"

Unknown

V_1

Solution

Rearrange the equation to isolate V_1. Plug in the known values and solve.

V_1=(C_2V_2)/C_1

V_1=(0.500color(red)cancel(color(black)("M"))xx"200.0 mL" )/(0.200color(red)cancel(color(black)("M")))= "50.0 mL" (rounded to three significant figures)

"50.0 mL" of the "2.00 M NaBr" solution is needed to make "200.0 mL" of a "0.500 M NaBr" solution.